Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 14

$$\eqalign{ & {\text{local maximum }}\left( { - \frac{2}{3},256} \right) \cr & {\text{local minimum }}\left( {2,0} \right) \cr & {\text{inflection point }}\left( {\frac{2}{3},128} \right) \cr & y{\text{ - intercept }}\left( {0,216} \right) \cr & x{\text{ - intercepts }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = 27{\left( {x - 2} \right)^2}\left( {x + 2} \right) \cr & {\text{Expanding}} \cr & f\left( x \right) = 27\left( {{x^2} - 4x + 4} \right)\left( {x + 2} \right) \cr & f\left( x \right) = 27{x^3} - 54{x^2} - 108x + 216 \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {27{x^3} - 54{x^2} - 108x + 216} \right] \cr & f'\left( x \right) = 81{x^2} - 108x - 108 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & 81{x^2} - 108x - 108 = 0 \cr & 3{x^2} - 4x - 4 = 0 \cr & \left( {x - 2} \right)\left( {3x + 2} \right) = 0 \cr & {\text{The critical points are }}x = 2{\text{ and }}x = - \frac{2}{3} \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {81{x^2} - 108x - 108} \right] \cr & f''\left( x \right) = 162x - 108 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = 2{\text{ and }}x = - \frac{2}{3} \cr & *f''\left( 2 \right) = 162\left( 2 \right) - 108 = 216 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = 2 \cr & f\left( 2 \right) = 27{\left( {2 - 2} \right)^2}\left( {2 + 2} \right) = 0 \cr & {\text{local minimum }}\left( {2,0} \right) \cr & *f''\left( { - \frac{2}{3}} \right) = 162\left( { - \frac{2}{3}} \right) - 108 = - 216 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}x = - \frac{2}{3} \cr & f\left( { - \frac{2}{3}} \right) = 27{\left( { - \frac{2}{3} - 2} \right)^2}\left( { - \frac{2}{3} + 2} \right) = 256 \cr & {\text{local maximum }}\left( { - \frac{2}{3},256} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0{\text{ to locate the inflection points}} \cr & f''\left( x \right) = 162x - 108 \cr & 162x - 108 = 0 \cr & x = \frac{2}{3} \cr & f\left( {\frac{2}{3}} \right) = 27{\left( {\frac{2}{3} - 2} \right)^2}\left( {\frac{2}{3} + 2} \right) = 128 \cr & {\text{The inflection point is }}\left( {\frac{2}{3},128} \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 27{\left( {0 - 2} \right)^2}\left( {0 + 2} \right) \cr & f\left( 0 \right) = 216 \cr & y{\text{ - intercept }}\left( {0,216} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 27{\left( {x - 2} \right)^2}\left( {x + 2} \right) = 0 \cr & x = - 2,{\text{ }}x = 2 \cr & x{\text{ - intercepts }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & \cr & {\text{Summary:}} \cr & {\text{local maximum }}\left( { - \frac{2}{3},256} \right) \cr & {\text{local minimum }}\left( {2,0} \right) \cr & {\text{inflection point }}\left( {\frac{2}{3},128} \right) \cr & y{\text{ - intercept }}\left( {0,216} \right) \cr & x{\text{ - intercepts }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & \cr & {\text{Graph}} \cr} $$